Homotopy between two closed curves

Question

I am confused about proving homotopy between closed curves that don't share endpoints.

The definition given in Stein/Shakarchi complex analysis is

Let $\gamma_{0}$ and $\gamma_{1}$ be two curves in an open set $\Omega$ with common end-points. So if $\gamma_{0}(t)$ and $\gamma_{1}(t)$ are two parametrizations defined on $[a, b]$, we have $$ \gamma_{0}(a)=\gamma_{1}(a)=\alpha \quad \text { and } \quad \gamma_{0}(b)=\gamma_{1}(b)=\beta $$ These two curves are said to be homotopic in $\Omega$ if for each $0 \leq s \leq 1$ there exists a curve $\gamma_{s} \subset \Omega$, parametrized by $\gamma_{s}(t)$ defined on $[a, b]$, such that for every $s$ $$ \gamma_{s}(a)=\alpha \quad \text { and } \quad \gamma_{s}(b)=\beta $$ and for all $t \in[a, b]$ $$ \left.\gamma_{s}(t)\right|_{s=0}=\gamma_{0}(t) \quad \text { and }\left.\quad \gamma_{s}(t)\right|_{s=1}=\gamma_{1}(t) $$ Moreover, $\gamma_{s}(t)$ should be jointly continuous in $s \in[0,1]$ and $t \in[a, b]$.

How can I show that one closed curve is homotopic to another if no endpoints are shared between the curves? Intuitively two nested circles should be homotopic but this would violate the endpoints being shared.

Answer 1

Let $f : g : X \to Y$ be maps between topological spaces $X,Y$. A homotopy from $f$ to $g$ is a continuous map $H : X \times [0,1] \to Y$ such that $H(x,0) = f(x)$ and $H(x,1) = g(x)$ for all $x \in X$. The family $\gamma_s$ is nothing else than a homotopy $H : [a,b] \times [0,1] \to \Omega$ from $\gamma_0$ to $\gamma_1$ with the property that the homotopy is stationary on the boundary $\{a, b\}$ of $[a,b]$. This means that $H(a,s) = \gamma_0(a) = \gamma_1(a)$ and $H(b,s) = \gamma_0(b) = \gamma_1(b)$ for all $s$. Let us call any such homotopy a curve homotopy.

Note that in other textbooks you frequently find the word path instead of curve.

You want to consider closed curves which we shall denote as loops. What you need is this concept:

Given loops $u,v$ in $X$, a loop homotopy is a homotopy $H : [a,b] \times [0,1] \to X$ from $u$ to $v$ such that for each $s \in [0,1]$ the map $H_s : [a,b] \to X, H_s(t) = H(t,s)$, is a loop. In that case we say that $u,v$ are loop homotopic, $u \simeq_l v$. Note that loop homotopies are in general not stationary on $\{a,b\}$; but they are not allowed to rip loops.

See also Characterizing simply connected spaces.